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An amount of ice of mass 10 –3 kg and temperature –10°C is transformed to vapour of temperature 110° by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100 Jkg –1 K –1 , specific heat of water = 4180 Jkg –1 K –1 , specific heat of steam = 1920 Jkg –1 K –1 , Latent heat of ice = 3.35 × 10 5 Jkg –1 and Latent heat of steam = 2.25 × 10 6 Jkg –1 )

1. 3022 J
2. 3043 J
3. 3003 J
4. 3024 J

Correct answer: 3043 J

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